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Inductance Consider 2 coils of wire Li = self inductance (how Ii affects itself) C2 Mij = mutual inductance C1 I1 The magnetic field through C1 is (how Ij affects Ii) I2 B1 I1 B2 I2 and similarly, the flux through C2 is This is important because When currents change, that will affect those and other currents by “inductance”. Theorem Proof M12 = M21 “Neumann’s equation” Q. E. D. Example. The mutual inductance between coaxial solenoids, Sol1 and Sol2, assuming these are densely wound solenoids and Sol2 is outside Sol1 ... (2) Assume l2 < l1 and A2 > A1. Neglect end effects. Note that M12 = M21. Self Inductance (L) LR Circuit Example: a solenoid (= a wire densely wound around a long cylinder) I = emf /R = - d/dt (LI) dI /dt = - (R/L) I I(t) = I0 exp(-Rt/L) Units: Tm/A x m2/m = Tm2/A = H (henry) /R Energy density of a magnetic field Apply energy conservation to the LR circuit. The power dissipated in resistance is P = I2 R (Joule’s law). Conservation of energy implies P dt = dQ V P = IV = I2R where U is the energy of the magnetic field in the inductor. So x